1. Physics 211: Lecture 5, Pg 1 Physics 211: Lecture 5 Today’s Agenda l More discussion of dynamics  Recap Newton's Laws Free Body Diagram  The Free Body Diagram  The tools we have for making & solving problems:  Ropes & Pulleys (tension)  Hooke’s Law (springs) Original: Mats A.Selen Modified: W.Geist

2. Physics 211: Lecture 5, Pg 2 Materials l Bring strings and springs. l Two spring powered carts on track: Use equal and unequal masses. Have students predict the outcome (i.e. accelerations and final speeds)

3. Physics 211: Lecture 5, Pg 3 Review: Newton's Laws Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Fa FF Law 2: For any object, F NET = ma Where F NET =  F action-reactionFF F Law 3: Forces occur in action-reaction pairs, F A,B = - F B,A. Where F A,B is the force acting on object A due to its interaction with object B and vice-versa. m is “mass” of object

4. Physics 211: Lecture 5, Pg 4 Gravity: Mass vs Weight l What is the force of gravity exerted by the earth on a typical physics student?  Typical student mass m = 55kg  g = 9.81 m/s 2.  F g = mg = (55 kg)x(9.81 m/s 2 )  F g = 540 N = WEIGHT F= -g F E,S = -mg F = F= g F S,E = F g = mg

5. Physics 211: Lecture 5, Pg 5 Lecture 5, Act 1 Mass vs. Weight l An astronaut on Earth kicks a bowling ball straight ahead and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon in the same manner with the same force. l His foot hurts... (a) more (b) less (c) the same

6. Physics 211: Lecture 5, Pg 6 Lecture 5, Act 1 Solution l The masses of both the bowling ball and the astronaut remain the same, so his foot will feel the same resistance and hurt the same as before. Ouch.

7. Physics 211: Lecture 5, Pg 7 Lecture 5, Act 1 Solution l However the weights of the bowling ball and the astronaut are less: l Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth. W = mg Moon g Moon < g Earth

8. Physics 211: Lecture 5, Pg 8 The Free Body Diagram Fa l Newton’s 2nd Law says that for an object F = ma. for an object. l Key phrase here is for an object.  Object has mass and experiences forces Fa l So before we can apply F = ma to any given object we isolate the forces acting on this object:

9. Physics 211: Lecture 5, Pg 9 The Free Body Diagram... l Consider the following case as an example of this….  What are the forces acting on the plank ?  Other forces act on F, W and E. focus on plank P = plank F = floor W = wall E = earth F F W,P F F P,W F F P,F F F P,E F F F,P F F E,P

10. Physics 211: Lecture 5, Pg 10 The Free Body Diagram... l Consider the following case  What are the forces acting on the plank ? Isolate the plank from the rest of the world. F F W,P F F P,W F F P,F F F P,E F F F,P F F E,P

11. Physics 211: Lecture 5, Pg 11 The Free Body Diagram... l The forces acting on the plank should reveal themselves... F F P,W F F P,F F F P,E

12. Physics 211: Lecture 5, Pg 12 Aside... l In this example the plank is not moving...  It is certainly not accelerating! Fa F  So F NET = ma becomes F NET = 0  This is the basic idea behind statics, which we will discuss in a few weeks. FFF F P,W + F P,F + F P,E = 0 F F P,W F F P,F F F P,E

13. Physics 211: Lecture 5, Pg 13 Example l Example dynamics problem: x A box of mass m = 2 kg slides on a horizontal frictionless floor. A force F x = 10 N pushes on it in the x direction. What is the acceleration of the box? Fi F = F x i a a = ? m y x

14. Physics 211: Lecture 5, Pg 14 Example... l Draw a picture showing all of the forces F F F B,F F F F,B F F B,E F F E,B y x

15. Physics 211: Lecture 5, Pg 15 Example... l Draw a picture showing all of the forces. l Isolate the forces acting on the block. F F F B,F F F F,B F g F B,E = mg F F E,B y x

16. Physics 211: Lecture 5, Pg 16 Example... l Draw a picture showing all of the forces. l Isolate the forces acting on the block. l Draw a free body diagram. F F F B,F gmggmg y x

17. Physics 211: Lecture 5, Pg 17 Example... l Draw a picture showing all of the forces. l Isolate the forces acting on the block. l Draw a free body diagram. l Solve Newton’s equations for each component.  F X = ma X  F B,F - mg = ma Y F F F B,F gmggmg y x

18. Physics 211: Lecture 5, Pg 18 Example... l F X = ma X  So a X = F X / m = (10 N)/(2 kg) = 5 m/s 2. l F B,F - mg = ma Y  But a Y = 0  So F B,F = mg. Normal Force l The vertical component of the force of the floor on the object (F B,F ) is often called the Normal Force (N). l Since a Y = 0, N = mg in this case. FXFX N mg y x

19. Physics 211: Lecture 5, Pg 19 Example Recap FXFX N = mg mg a X = F X / m y x

20. Physics 211: Lecture 5, Pg 20 Lecture 5, Act 2 Normal Force (Check your answer: Similar to mass attached to string) l A block of mass m rests on the floor of an elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block? m (a) N > mg (b) N = mg (c) N < mg a

21. Physics 211: Lecture 5, Pg 21 Lecture 5, Act 2 Solution m N mg All forces are acting in the y direction, so use: F total = ma N - mg = ma N = ma + mg therefore N > mg a

22. Physics 211: Lecture 5, Pg 22 Tools: Ropes & Strings l Can be used to pull from a distance. l Tension l Tension (T) at a certain position in a rope is the magnitude of the force acting across a cross-section of the rope at that position.  The force you would feel if you cut the rope and grabbed the ends.  An action-reaction pair. cut T T T Tension doesn’t have a direction. When you hook up a wire to an object the direction is determined by geometry of the hook up.

23. Physics 211: Lecture 5, Pg 23 Tools: Ropes & Strings... l Consider a horizontal segment of rope having mass m:  Draw a free-body diagram (ignore gravity). x l Using Newton’s 2nd law (in x direction): F NET = T 2 - T 1 = ma So if m = 0 (i.e. the rope is light) then T 1 =  T 2  T is constant anywhere in a rope or string as long as its massless T1T1 T2T2 m ax

24. Physics 211: Lecture 5, Pg 24 Tools: Ropes & Strings... l An ideal (massless) rope has constant tension along the rope. l If a rope has mass, the tension can vary along the rope  For example, a heavy rope hanging from the ceiling... l We will deal mostly with ideal massless ropes. T = T g T = 0 TT 2 skateboards

25. Physics 211: Lecture 5, Pg 25 Tools: Ropes & Strings... l What is force acting on box by the rope in the picture below? (always assume rope is massless unless told different) mg T m Since a y = 0 (box not moving), T = mg

26. Physics 211: Lecture 5, Pg 26 Lecture 5, Act 3 Force and acceleration l A fish is being yanked upward out of the water using a fishing line that breaks when the tension reaches 180 N. The string snaps when the acceleration of the fish is observed to be is 12.2 m/s 2. What is the mass of the fish? m = ? a = 12.2 m/s 2 snap ! (a) 14.8 kg (b) 18.4 kg (c) 8.2 kg For what object do you draw the FBD?

27. Physics 211: Lecture 5, Pg 27 Lecture 5, Act 3 Solution: l Draw a Free Body Diagram!! T mg m = ? a = 12.2 m/s 2 l Use Newton’s 2nd law in the upward direction: F TOT = ma T - mg = ma T = ma + mg = m(g+a)

28. Physics 211: Lecture 5, Pg 28 Tools: Pegs & Pulleys l Used to change the direction of forces  An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: FF1FF1 ideal peg or pulley FF2FF2 FF | F 1 | = | F 2 |

29. Physics 211: Lecture 5, Pg 29 Tools: Pegs & Pulleys l Used to change the direction of forces  An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: mg T m T = mg F W,S = mg

30. Physics 211: Lecture 5, Pg 30 Springs l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.  F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position F X = 0 x

31. Physics 211: Lecture 5, Pg 31 Springs... l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.  F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position F X = -kx > 0 x x  0

32. Physics 211: Lecture 5, Pg 32 Springs... l Hooke’s Law: l Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.  F X = -k xWhere x is the displacement from the relaxed position and k is the constant of proportionality. F X = - kx < 0 x x > 0 relaxed position Horizontal springs

33. Physics 211: Lecture 5, Pg 33 Scales: l Springs can be calibrated to tell us the applied force.  We can calibrate scales to read Newtons, or...  Fishing scales usually read weight in kg or lbs. 0 2 4 6 8 1 lb = 4.45 N Spring/string

34. Physics 211: Lecture 5, Pg 34 mmm (a) (b) (c) (a) 0 lbs. (b) 4 lbs. (c) 8 lbs. (1) (2) ? Lecture 5, Act 4 Force and acceleration l A block weighing 4 lbs is hung from a rope attached to a scale. The scale is then attached to a wall and reads 4 lbs. What will the scale read when it is instead attached to another block weighing 4 lbs? Scale on a skate

35. Physics 211: Lecture 5, Pg 35 Lecture 5, Act 4 Solution: l Draw a Free Body Diagram of one of the blocks!! l Use Newton’s 2nd Law in the y direction: F TOT = 0 T - mg = 0 T = mg = 4 lbs. mg T m T = mg a = 0 since the blocks are stationary

36. Physics 211: Lecture 5, Pg 36 Lecture 5, Act 4 Solution: l The scale reads the tension in the rope, which is T = 4 lbs in both cases! mmm T T T T T T T

37. Physics 211: Lecture 5, Pg 37 Recap of today’s lecture.. l More discussion of dynamics (Read Chapter 4 in text) èRecap Free Body Diagram èThe Free Body Diagram èThe tools we have for making & solving problems: »Ropes & Pulleys (tension) »Hooke’s Law (springs).